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QUESTION

Let $U_n$ denote the group of units modulo $n$.

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\item[(i)]
Explain the following terms: (a) $g\in U_n$ is a primitive root
and (b) $g\in U_n$ is a quadratic non-residue.

\item[(ii)]
Suppose that $p=2^m+1$ is a prime for some $m>0$. Show rhat $g\in
U_p$ is a quadratic non-residue if and only if it is a primitive
root.

\item[(iii)]
Using quadratic reciprocity, or otherwise, show that

$$3^{\frac{(p-1)}{2}}\equiv-1(\textrm{modulo }p).$$

\item[(iv)]
If $n$ is an integer of the form $n=2^{2^m}+1$ such that
$3^{\frac{(n-1)}{2}}\equiv-1$ (modulo $n$) show that $n$ is a
prime. (Hint: In the proof of (iv) you may assume (see question
8(viii)) without proof that $\phi(n)\leq n-\sqrt{n}$ if $p$ is a
composite integer.)

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ANSWER

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\item[(i)]
If $U_n$ is a cyclic group then any generator, $g\in U_n$ is
called a primitive root modulo $n$. A quadratic non-residue, $g\in
U_n$, is any element for which the equation $g=h^2$ has no
solution $f\in U_n$.

\item[(ii)]
If $p=2^{2^m}+1$ is a prime then $U_p$ is cyclic of order
$p-1=2^{2^m}$. Choose a generator, $g\in U_p$. Then $g^j$ is a
generator if and only if gcd$(j,2^{2^m})=1$ is odd, which is
equivalent to $j$ being odd. On the other hand $g^j=h^2=g^{2s}$
has a solution if and only if $j$ is even. Hence $g^j$ is a
quadratic non-residue if and only if $j$ is odd.

\item[(iii)]
If the order of 3 in $U_p$ is equal to $2^\alpha$ then
$3^{2^{\alpha-1}}=-1$, since $3^{2^{\alpha-1}}$ is not congruent
to 1 (modulo $p$) but its square is. Hence the given congruence is
equivalent, by part (ii), to 3 being a quadratic non-residue
modulo $p$. In terms of Legendre symbols

$$\left(\frac{3}{P}\right)=-1$$

if and only if

$$3^{\frac{(p-1)}{2}}\equiv-1\textrm{ (modulo }p).$$

By Quadratis Reciprocity

$$\left(\frac{3}{P}\right)\left(\frac{p}{3}\right)=(-1)^{(s-1)(p-1)/4}=1$$

However $p\equiv(-1)^{2^m}+1\equiv2$ (modulo 3) and

$$\left(\frac{2}{3}\right)=-1,$$

as required.

\item[(iv)]
If $3^{\frac{(n-1)}{2}}\equiv-1$ (modulo $n$) then the order of 3
in $U_n$ is at least $n-1$. However, $g^{\phi(n)}\equiv1$ for all
$g\in U_n$. Hence if $n$ is composite the Hint yields

$$n-1\leq\phi(n)\leq n-\sqrt{n}$$

which is impossible. Hence $n$ must be prime.

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